Question
A space vehicle moving in a circular orbit of radius $r_{1}$ transfers to a larger circular orbit of radius $r_{2}$ by means of an elliptical path between $A$ and $B$. (This transfer path is known as the Hohmann transfer ellipse.) The transfer is accomplished by a burst of speed $\Delta v_{A}$ at $A$ and a second burst of speed $\Delta v_{B}$ at $B$ Write expressions for $\Delta v_{A}$ and $\Delta v_{B}$ in terms of the radii shown and the value of $g$ of the acceleration due to gravity at the earth's surface. If each $\Delta v$ is positive, how can the velocity for path 2 be less than the velocity for path $1 ?$ Compute each $\Delta v$ if $r_{1}=(6371+$ 500) $\mathrm{km}$ and $r_{2}=(6371+35800) \mathrm{km} .$ Note that $r_{2}$ has been chosen as the radius of a geosynchronous orbit.
Step 1
The velocity in a circular orbit is given by the formula $v = \sqrt{\frac{G}{r}}$, where $G$ is the gravitational constant and $r$ is the radius of the orbit. So, we have: \[v_1 = \sqrt{\frac{G}{r_1}}\] \[v_2 = \sqrt{\frac{G}{r_2}}\] Show more…
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A spacecraft of mass $m$ describes a circular orbit of radius $r_{1}$ around the earth. $(a)$ Show that the additional energy $\Delta E$ that must be imparted to the spacecraft to transfer it to a circular orbit of larger radius $r_{2}$ is $$\Delta E=\frac{G M m\left(r_{2}-r_{1}\right)}{2 r_{1} r_{2}}$$ where $M$ is the mass of the earth. ( $b$ ) Further show that if the transfer from one circular orbit to the other is executed by placing the spacecraft on a transitional semielliptic path $A B,$ the amounts of energy $\Delta E_{A}$ and $\Delta E_{B}$ which must be imparted at $A$ and $B$ are, respectively, proportional to $r_{2}$ and $r_{1}$ : $$\Delta E_{A}=\frac{r_{2}}{r_{1}+r_{2}} \Delta E \quad \Delta E_{B}=\frac{r_{1}}{r_{1}+r_{2}} \Delta E$$
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